Find an equation of the line that is tangent to the graph of f and parallel to the given line.

Function Line
f(x) = 2x^2 6x – y + 4 = 0

I know i have to find derivative 1st which will be f'(x)=4x but then how to i find equation do i plug 6 in for x so 4(6) would it be 6x-24???

To find an equation of a line that is tangent to the graph of function f and parallel to the given line, you need to follow these steps:

1. Find the derivative of function f: f'(x).
In this case, the given function is f(x) = 2x^2 + 6x – y + 4 = 0. To find the derivative, differentiate each term with respect to x while treating y as a constant:
f'(x) = d/dx(2x^2) + d/dx(6x) - d/dx(y) + d/dx(4)
= 4x + 6 - 0 + 0
= 4x + 6

2. Since you want to find a line parallel to the given line, the slope of the tangent line and the given line need to be the same.
The given line has a slope of 6, so the slope of the tangent line should also be 6.

3. Set the derivative of f equal to the slope (6) and solve for x.
4x + 6 = 6
4x = 6 - 6
4x = 0
x = 0

4. Plug the x-value found (x = 0) into the original function f to find the corresponding y-value.
f(0) = 2(0)^2 + 6(0) - y + 4 = 0
0 + 0 - y + 4 = 0
-y + 4 = 0
-y = -4
y = 4

So, the point at which the tangent line touches the graph of f is (0, 4).

5. Use the point-slope form of a linear equation (y - y1 = m(x - x1)) to write the equation of the tangent line, where (x1, y1) is the point we found and m is the slope.
Plugging in the values, we have:
y - 4 = 6(x - 0)
y = 6x + 4

Therefore, the equation of the line that is tangent to the graph of f and parallel to the given line is y = 6x + 4.